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Consider a simpler algorithm:

• Pass 1: Build a map of counts of characters. You can use a simple int array to store this mapping.
• Pass 2: For each index, if the count of the current character is 1, return the index.
• If pass 2 terminates without returning, then there are no unique characters, return -1.

This is simple, straightforward logic. The constant number of passes doesn't change the time complexity.

You could do it in a single pass too, for example using a queue to keep track of the ordering of what was seen first, but I would prefer to avoid the added complexity unless there is a compelling reason.